Q.A displacement vector of magnitude 10 m makes an angle of 30∘ with the positive x-axis. Find its rectangular components Ax and Ay.
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Concept understanding — Resolution of Vectors
Just as several vectors can be combined into one resultant, a single vector can conversely be split into components along chosen directions: V=V1n^1+V2n^2+V3n^3, where V1,V2,V3 are the components along the unit directions n^1,n^2,n^3. This splitting process is called resolution. When the chosen directions are mutually perpendicular, the components are called rectangular components — by far the most useful and common choice.
For a vector R in a two-dimensional rectangular coordinate system making angle θ with the x-axis, the rectangular components are Rx=Rcosθ and Ry=Rsinθ, so that R=Rxi^+Ryj^. From these, the magnitude is recovered as R=Rx2+Ry2 and the direction as θ=tan−1(Ry/Rx); the same idea extends naturally to three dimensions with a third component Rz along k^, giving R=Rx2+Ry2+Rz2.
An important consequence of resolution is the component test for vector equality: two vectors are equal if and only if ALL their corresponding rectangular components are equal (Ax=Bx, Ay=By, Az=Bz). The same logic extends to testing whether two vectors are PARALLEL: they are parallel exactly when one vector's components are a constant multiple of the other's, i.e. when Ax/Bx=Ay/By=Az/Bz for some fixed scalar ratio.
[!TLDR] Use Ax=Acosθ, Ay=Asinθ. [!ANSWER] Ax=53≈8.66 m and Ay=5 m.
With A=10 m and θ=30∘: Ax=Acosθ=10cos30∘=10×23=53≈8.66 m, and Ay=Asinθ=10sin30∘=10×0.5=5 m. As a check, Ax2+Ay2=75+25=100=10 m, matching the original magnitude. [!ANSWER] Ax=53≈8.66 m and Ay=5 m.
Directly apply the rectangular-resolution formulas Ax=Acosθ, Ay=Asinθ, then check with the Pythagorean relation.
Swapping sine and cosine (using sinθ for the x-component).\n- Using degrees/radians inconsistently in the calculator.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
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Q.A boy pulls a string attached to a wooden block with a force of 30 N. The angle between the string and the horizontal ground is 30°. How much is
(a) the effective value of the force that tends to move the block along the ground
(b) the quantity of force tending to lift the block vertically upward.
›Reveal solutionSolution
Resolving the 30 N force at 30 degrees to the horizontal into its horizontal (F cos 30°) and vertical (F sin 30°) components gives about 26 N moving the block and 15 N tending to lift it.
Given: applied force F = 30 N, angle with horizontal ground theta = 30 degrees
Horizontal component (this is the effective force that drags/moves the block ALONG the ground):
F_horizontal = F cos(theta) = 30 x cos(30°) = 30 x (sqrt(3)/2) = 30 x 0.866 ≈ 25.98 N (≈ 26 N)
Vertical component (this is the force that tends to LIFT the block off the ground, reducing its effective weight/normal reaction):
F_vertical = F sin(theta) = 30 x sin(30°) = 30 x 0.5 = 15 N
✓Final answer
The effective horizontal force moving the block along the ground is about 25.98 N (≈ 26 N).
The vertical component tending to lift the block is 15 N.